1.CH3-p4 Is o(dt) meaning 0? I never saw it before. 2.CH3-p18-Section 4.3 Asymptotic distribution of mu Why do we discuss the asymptotic distribution of mu in particular? What is the meaning of this step? What will happen if mu is not a asymptotic distribution? 3.CH3-p31 Why does the first term in the integrating factor method disappear? I don't know how to integrate the first term.It's strange. 4.CH4-p11 In Proof, why P[N_(t+s)=1|T_0=s]=P[N_(t+s)-N_s=0|T_0=s]? I can understand that T_1 is the probability that 1 events occur between time 0 and time t+s. Since T_0 is the probability that no events occur between time 0 and time s, doesn't it mean N_s=0?
Hi Ykai 1. the definition of an o(t) function is in the next paragraph of the notes. 2. Firstly it's worth remembering some of the notation. Mu is the true unknown transition rate, it is not random nor does it have a distribution. This section looks at the asymptotic distribution of mu with a tilde over the top, which is how we denote the estimator of mu. The estimator is a random variable and so has a distribution. We don't have a result for the exact distribution for a finite sample size but we do have the asymptotic distribution, which is derived in this section. So we can use this distribution to, for example, (approximately) construct confidence intervals and do hypothesis tests. 3. I'm not sure exactly what step you're referring to but I going to guess it's the integration of (**). The first term here doesn't disappear so to speak. To check that the integration as stated works, the easiest thing to do is use the chain rule to do the differentiation (ie differentiate what we claim to be the integral of the LHS of (**)). We have: \( y e^{\int{P(x)} dx} \) Using the chain rule, the derivative of this is: \( dy/dx \) \( e^{\int{P(x)} dx} + y \frac{d}{dx} e^{\int{P(x)} dx}\) This is: \( dy/dx \) \( e^{\int{P(x)} dx} + y P(x) e^{\int{P(x)} dx}\) As per the LHS of (**). If you do the preceding steps of the integrating factor method correctly, the two terms on the LHS will always integrate to y * IF where y is the function of interested and IF is the integrating factor. 4. T0 is the amount of time spent in state 0 - alternatively it is the amount of time until the first event. P[N_(t+s)=1|T_0=s] is the probability that one event has happened up to time t+s given that the first event happened at time s. This means we can have no more events happen between times s and s +t. In other words N_(s) = N_(s+t) in this scenario and so the probability is the same as P[N_(t+s)-N_s=0|T_0=s]. Hope this helps! Andy
